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MATH582课程简介
Overview
Mathematics & Statistics (Sci) : CW-complexes, cellular approximation theorem. Homotopy groups, long exact sequence for a fiber bundle. Whitehead theorem. Freudenthal suspension theorem. Singular and cellular homology and cohomology. Hurewicz theorem. Mayer-Vietoris sequence. Universal coefficients theorem. Cup product, Kunneth formula, Poincare duality.
CW-complexes are a type of topological space that are built up inductively from points, lines, disks, and higher-dimensional analogues. The construction proceeds by attaching cells of increasing dimension, following a certain attaching map. The resulting space is called a CW-complex, and it has some very nice properties that make it a useful tool in algebraic topology.
Cellular Approximation Theorem: The cellular approximation theorem says that any continuous map from a CW-complex X to another space Y can be approximated arbitrarily well by a cellular map. This means that we can study the topology of X by studying cellular maps from X to simpler spaces, like spheres or tori.
Prerequisites
Terms: Winter 2023
Instructors: Przytycki, Piotr (Winter)
Prerequisite: MATH 576 or equivalent or permission of instructor.
Course to be taught alternate years – Winter Semester
Homotopy Groups: The n-th homotopy group of a space X, denoted by π_n(X), measures the ways in which loops in X can be continuously deformed to one another. It is a group, meaning that there is a way to combine two loops to get another loop, and there is an identity loop that does nothing. Homotopy groups are a powerful tool for studying the structure of spaces.
Long Exact Sequence for a Fiber Bundle: A fiber bundle is a space that locally looks like a product space, but globally has a more complicated structure. The long exact sequence for a fiber bundle relates the homotopy groups of the total space, the base space, and the fiber, and gives a way to compute them in terms of each other.
MATH582 Topology HELP（EXAM HELP， ONLINE TUTOR）
Throughout, a topological space $X$ is endowed with a topology $\tau$, even if not explicitly mentioned.
A collection $\left{A_\alpha\right}$ of subsets of $X$ satisfies the finite intersection property if for any finite subcollection.
(a) Prove Cantor’s “finite intersection lemma”: Suppose $\left{K_\alpha\right}$ is a collection of compact sets of a Hausdorff space $X$. If $\bigcap_{i=1}^n K_{\alpha_i} \neq \emptyset$ for any finite subcollection, then $\bigcap_\alpha K_\alpha \neq$ $\emptyset$.
(b) Prove that $X$ is compact if and only if every collection of closed sets $\left{F_\alpha\right}$ satisfying the finite intersection property must also satify $\bigcap_{\alpha \in I} F_\alpha \neq \emptyset$.
(a) Suppose for contradiction that $\bigcap_\alpha K_\alpha = \emptyset$. Then for each $\alpha$, there exists an open set $U_\alpha$ containing $K_\alpha$ such that $U_\alpha \cap \left(\bigcap_{\beta \neq \alpha} K_\beta\right) = \emptyset$. This follows from the fact that $X$ is Hausdorff and for any $x \in K_\alpha$, there exist disjoint open sets $U_x$ and $V_\alpha$ containing $x$ and $K_\alpha$ respectively, and we can take $U_\alpha = \bigcap_{x \in K_\alpha} U_x$.
Then $\left{U_\alpha\right}$ is an open cover of $\bigcup_\alpha K_\alpha$. Since $\left{K_\alpha\right}$ is a collection of compact sets, there exists a finite subcover $\left{U_{\alpha_1}, \dots, U_{\alpha_n}\right}$ of $\bigcup_{i=1}^n K_{\alpha_i}$. But then $\bigcap_{i=1}^n K_{\alpha_i} \subseteq \bigcap_{i=1}^n U_{\alpha_i} \cap \left(\bigcap_{\beta \neq \alpha_i} K_\beta\right) = \emptyset$, which contradicts the assumption that $\bigcap_{i=1}^n K_{\alpha_i} \neq \emptyset$ for any finite subcollection.
(b) $(\Rightarrow)$ Suppose $X$ is compact and let $\left{F_\alpha\right}$ be a collection of closed sets satisfying the finite intersection property. Then for each finite subcollection $\left{F_{\alpha_1}, \dots, F_{\alpha_n}\right}$, we have $\bigcap_{i=1}^n F_{\alpha_i} \neq \emptyset$. Let $G_i = X \setminus F_{\alpha_i}$ for each $i$. Then $\left{G_i\right}$ is a collection of open sets such that $\bigcap_{i=1}^n G_i = X \setminus \bigcap_{i=1}^n F_{\alpha_i} = X \setminus \emptyset = X$. By compactness of $X$, there exists a finite subcover $\left{G_{i_1}, \dots, G_{i_m}\right}$ of $\left{G_i\right}$. Then $\bigcap_{j=1}^m F_{\alpha_{i_j}} = X \setminus \bigcup_{j=1}^m G_{i_j} = X \setminus X = \emptyset$. Therefore, $\left{F_\alpha\right}$ satisfies $\bigcap_{\alpha \in I} F_\alpha \neq \emptyset$.
$(\Leftarrow)$ Suppose that every collection of closed sets $\left{F_\alpha\right}$ satisfying the finite intersection property must also satisfy $\bigcap_{\alpha \in I} F_\alpha \neq \emptyset$. Let $\left{U_\alpha\right}$ be an open cover of $X$. For each $\alpha$, let $F_\alpha = X \setminus U_\alpha$. Then $\left{F_\alpha\right}$ is a collection of closed sets and $\bigcap_{\alpha} U_\alpha =
Let $d \rho$ be a probability measure on $\mathbb{R}$ with $\int x d \mu(x)=0$ and $\int x^4 d \rho(x)<\infty$. Prove the central limit theorem for the sum of independent random variables with this distribution.
Specifically, if $X_1, X_2, \ldots$ are $d \rho$-distributed, show that for any Schwartz function $f$
$$
\mathbb{E}\left{f\left(\frac{X_1+\cdots+X_n}{n^{1 / 2}}\right)\right} \longrightarrow \frac{1}{\sqrt{2 \pi \sigma^2}} \int \exp \left{-\frac{x^2}{2 \sigma^2}\right} f(x) d x
$$
as $n \rightarrow \infty$. Hint: first show convergence for $f(x)=e^{-2 \pi i x \xi}$ uniformly for $\xi$ in a compact set.
(a) Suppose $f$ is continuous and let $(x,y) \in X \times Y$ be a limit point of $\Gamma_f$. This means that there exists a sequence ${(x_n, f(x_n))}$ in $\Gamma_f$ such that $(x_n, f(x_n)) \to (x,y)$ as $n \to \infty$. By definition of the product topology, this implies that $x_n \to x$ in $X$ and $f(x_n) \to y$ in $Y$.
Since $Y$ is Hausdorff, the limit of a sequence in $Y$ is unique, so $f(x_n) \to y$ and $f(x) = y$. Therefore, $(x,y) \in \Gamma_f$, and we have shown that $\Gamma_f$ contains all its limit points, so it is closed.
(b) Let $X = {a,b}$ with the discrete topology, and let $Y$ be any topological space that is not Hausdorff. Define $f: X \to Y$ by $f(a) = f(b)$. Then $\Gamma_f = {(a,f(a)), (b,f(b))} = {(a,y)} \cup {(b,y)}$, which is the union of two copies of the point ${a,b}$ in $X$ and $f(a) = f(b)$ in $Y$. This set is not closed in $X \times Y$, since any open set containing $(a,f(b))$ or $(b,f(a))$ intersects $\Gamma_f$.
(c) Suppose $X$ and $Y$ are both compact and Hausdorff, and let $\Gamma_f$ be closed in $X \times Y$. To show that $f$ is continuous, let $U$ be an open subset of $Y$. Then $Y \setminus U$ is closed in $Y$, so $V = \Gamma_f \cap (X \times (Y \setminus U))$ is closed in $\Gamma_f$ (since the projection of $\Gamma_f$ onto the second coordinate is $Y$, and $Y \setminus U$ is closed in $Y$). Therefore, $V$ is compact (since $\Gamma_f$ is compact as a closed subset of the compact space $X \times Y$).
Now, let $x \in f^{-1}(U)$. Then $(x, f(x)) \in \Gamma_f$ and $f(x) \in U$, so $(x, f(x)) \notin X \times (Y \setminus U)$. Therefore, $(x, f(x)) \notin V$, so there exists an open neighborhood $W_x \times U_x$ of $(x, f(x))$ in $X \times Y$ such that $W_x \times U_x \cap V = \emptyset$. Then $W_x$ is an open neighborhood of $x$ in $X$ that maps into $U$ under $f$, since $f(W_x) \subseteq U_x \subseteq U$.
Since $X$ is compact, there exists a finite subcover ${W_{x_1}, \dots, W_{x_n}}$ of an open cover of $f^{-1}(U)$. Let $W = \bigcap_{i=1}^n W_{x_i}$, which is an open neighborhood of $f^{-1}(U)$ in $X$. Then $f(W) \subseteq \bigcup_{i=1}^n U_{x_i} \subseteq U$, so $
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